# Difference between revisions of "2008 AMC 10B Problems/Problem 10"

## Problem

Suppose that $(u_n)$ is a serquence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$,

and that $u_3=9$ and $u_6=128$. What is $u_5$?

(A) 40 (B) 53 (C) 68 (D) 88 (E) 104

## Solution

We know that $u_6=128$, so we plug in $n=4$ to get $128=2u_5+u_4$. We plug in $n=3$ to get

$u_5=2u_4+9$. Substituting gives

$128=5u_4+18 \rightarrow u_4=22$

This gives $u_5=\frac{128-22}{2}=53$.